Denoting the largest possible value of $N$ by $N(p,n)$, one has $N(p,1)=p-1$. In general, $N(p,n)\ge(p-1)n$, as follows by taking $(v_j)$ to be the sequence containing $p-1$ instances of every basis vector; how sharp is this bound?

Incidentally, ($\ast$) implies that for any subspace $L\le\mathbb F_p^n$ there are at most $|L|-1$ indices $j\in[1,N]$ with $v_j\in L$, but this does not lead to any reasonable upper bound for $N(p,n)$.

$\begingroup$The obvious upper bound is $np\log p$ (every next choice of $f_j$ can be made to reduce the available set of $z$ by $1/p$ of its size). I'm not sure which of the two trivial bounds is closer to the truth though...$\endgroup$
– fedjaNov 22 '18 at 19:21